LightOJ - 1079 Just another Robbery(背包)

题目大意：给出一个概率P，和N个银行，再给出每个银行的被捕概率和钱的数量，问在被捕概率小于P的情况下，偷到的最多的钱

解题思路：用dp[i][j]表示前i个银行，偷盗了j的钱的最低被捕概率，接着进行背包

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110;
const int M = 10010;
const double esp = 1e-6;

double p;
int n, sum, cas = 1;
int money[N];
double caught[N], dp[N][M];

void init() {
    scanf("%lf%d", &p, &n);
    sum = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d%lf", &money[i], &caught[i]);
        sum += money[i];
    }
}

void solve() {
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;

    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= sum; j++) {
            //第i个银行不抢
            dp[i][j] = dp[i - 1][j];
            //抢
            if (j - money[i] >= 0 && (double)1 + dp[i - 1][j - money[i]] > esp) {
                if ((double)1 + dp[i][j] > esp) 
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - money[i]] + (1 - dp[i - 1][j - money[i]]) * caught[i]);
                else 
                    dp[i][j] = dp[i - 1][j - money[i]] + (1 - dp[i - 1][j - money[i]]) * caught[i];
            }
        }
    for (int i = sum; i >= 0; i--)
        if (p - dp[n][i] > esp) {
            printf("Case %d: %d\n", cas++, i);
            break;
        }
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}